Question

find the oryhocentre of the triangle formed by vertices (5,-2),(-1,2),and(1,4)

pls ans me if u know it's very important​

Answer 1

Let A = (5, -2), B = (- 1, 2) and C = (1, 4)

Equation of altitude from C to AB : - 3x + 2y = 5 …….(1)

Equation of altitude from A to BC: x + y = 3 ……..(2)

Solving (1) and (2)

-3x + 2y = 5………….(1) x 1

2x + 2y = 6 ……………. (2) x 2

-5x = - 1

x = 1/5

= 0.2

0.2 + y = 3

y = 2.8

Orthocentre = (0.2, 2.8)

Answer 2

Let vertex (5,-2) be A, (-1,2) be B and (1,4) be C. Now let's find slopes of AB, AC and BC

mAB = (-2-(-1))/(5-2), mAB = - 1/3

mAC = (-2 -4)/(5–1), mAC = -3/2

mBC = (4-2)/(1-(-1)), mBC = 1

So after finding slopes of these line segments, we can make equation of perpendicular of these segments by using reciprocal of BC and coordinates A, reciprocal of AC and coordinates B.

Y = mX + b, - 2 = -1 × 5 + b, b = 3

(E)BC = Y = -X + 3..(1)

2 = 2/3×(-1) + b, b = 2 + 2/3, b = 8/3

(E)AC = Y = 2X/3 + 8/3..(2)

Subtracting first equation from second equation, we can get coordinates of intersection point of these two perpendicular lines, which is the orthocentre of triangle ABC.

2X/3+ X + 8/3 -3 = 0, 5X/3 = 3 - 8/3, 5X/3 = 1/3 , X = 1/5

Y = - 1/5 + 3, Y = (-1+15)/5, Y = 14/5

So the orthocentre of triangle is (1/5,14/5) or

(0.2,2.8)